Problem: $\dfrac{ -b + 8c }{ -2 } = \dfrac{ -3b + 9d }{ -10 }$ Solve for $b$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ -b + 8c }{ -{2} } = \dfrac{ -3b + 9d }{ -10 }$ $-{2} \cdot \dfrac{ -b + 8c }{ -{2} } = -{2} \cdot \dfrac{ -3b + 9d }{ -10 }$ $-b + 8c = -{2} \cdot \dfrac { -3b + 9d }{ -10 }$ Multiply both sides by the right denominator. $-b + 8c = -2 \cdot \dfrac{ -3b + 9d }{ -{10} }$ $-{10} \cdot \left( -b + 8c \right) = -{10} \cdot -2 \cdot \dfrac{ -3b + 9d }{ -{10} }$ $-{10} \cdot \left( -b + 8c \right) = -2 \cdot \left( -3b + 9d \right)$ Distribute both sides $-{10} \cdot \left( -b + 8c \right) = -{2} \cdot \left( -3b + 9d \right)$ ${10}b - {80}c = {6}b - {18}d$ Combine $b$ terms on the left. ${10b} - 80c = {6b} - 18d$ ${4b} - 80c = -18d$ Move the $c$ term to the right. $4b - {80c} = -18d$ $4b = -18d + {80c}$ Isolate $b$ by dividing both sides by its coefficient. ${4}b = -18d + 80c$ $b = \dfrac{ -18d + 80c }{ {4} }$ All of these terms are divisible by $2$ $b = \dfrac{ -{9}d + {40}c }{ {2} }$